Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct node {
  int v;
  int next;
};
node edge[200010];
vector<int> per[100010][20];
vector<int> pre[100010];
vector<int> ans;
int dp[100010][20];
int first[100010], deep[100010];
int n, m, num;
void addedge(int u, int v) {
  edge[num].v = v;
  edge[num].next = first[u];
  first[u] = num++;
  return;
}
void dfs(int cur, int pre) {
  int i, v;
  for (i = first[cur]; i != -1; i = edge[i].next) {
    v = edge[i].v;
    if (v != pre) {
      dp[v][0] = cur;
      deep[v] = deep[cur] + 1;
      dfs(v, cur);
    }
  }
  return;
}
void init() {
  vector<int>::iterator iter;
  int i, j, k, t;
  for (i = 1; i <= n; i++) {
    sort(pre[i].begin(), pre[i].end());
    pre[i].erase(unique(pre[i].begin(), pre[i].end()), pre[i].end());
  }
  for (i = 1; i <= n; i++) {
    for (k = 0; k < pre[i].size(); k++) {
      per[i][0].push_back(pre[i][k]);
    }
    for (k = 0; k < pre[dp[i][0]].size(); k++) {
      per[i][0].push_back(pre[dp[i][0]][k]);
    }
    sort(per[i][0].begin(), per[i][0].end());
    per[i][0].erase(unique(per[i][0].begin(), per[i][0].end()),
                    per[i][0].end());
    iter = per[i][0].begin(), t = per[i][0].size();
    for (k = 0; k < min(t, 10); k++) iter++;
    per[i][0].erase(iter, per[i][0].end());
  }
  for (j = 1; j <= log2(n); j++) {
    for (i = 1; i <= n; i++) {
      dp[i][j] = dp[dp[i][j - 1]][j - 1];
      for (k = 0; k < per[i][j - 1].size(); k++) {
        per[i][j].push_back(per[i][j - 1][k]);
      }
      for (k = 0; k < per[dp[i][j - 1]][j - 1].size(); k++) {
        per[i][j].push_back(per[dp[i][j - 1]][j - 1][k]);
      }
      sort(per[i][j].begin(), per[i][j].end());
      per[i][j].erase(unique(per[i][j].begin(), per[i][j].end()),
                      per[i][j].end());
      iter = per[i][j].begin(), t = per[i][j].size();
      for (k = 0; k < min(t, 10); k++) iter++;
      per[i][j].erase(iter, per[i][j].end());
    }
  }
  return;
}
int getlca(int u, int v) {
  int i;
  if (deep[u] < deep[v]) swap(u, v);
  for (i = log2(n); i >= 0; i--) {
    if (deep[dp[u][i]] >= deep[v]) {
      u = dp[u][i];
    }
  }
  if (u == v) return u;
  for (i = log2(n); i >= 0; i--) {
    if (dp[u][i] != dp[v][i]) {
      u = dp[u][i], v = dp[v][i];
    }
  }
  return dp[u][0];
}
void solve(int u, int lca) {
  int i, j;
  for (i = log2(n); i >= 0; i--) {
    if (deep[dp[u][i]] >= deep[lca]) {
      for (j = 0; j < per[u][i].size(); j++) {
        ans.push_back(per[u][i][j]);
      }
      u = dp[u][i];
    }
  }
  return;
}
int main() {
  int q, i, u, v, c, t, lca;
  scanf("%d%d%d", &n, &m, &q);
  memset(first, -1, sizeof(first));
  num = 0;
  for (i = 1; i <= n - 1; i++) {
    scanf("%d%d", &u, &v);
    addedge(u, v);
    addedge(v, u);
  }
  for (i = 1; i <= m; i++) {
    scanf("%d", &c);
    if (pre[c].size() < 10) {
      pre[c].push_back(i);
    }
  }
  deep[1] = 1;
  dfs(1, 0);
  init();
  while (q--) {
    scanf("%d%d%d", &u, &v, &c);
    lca = getlca(u, v);
    ans.clear();
    if (u == v) {
      t = pre[u].size();
      cout << min(t, c) << " ";
      for (i = 0; i < min(t, c); i++) {
        cout << pre[u][i] << " ";
      }
      cout << endl;
    } else {
      solve(u, lca);
      solve(v, lca);
      sort(ans.begin(), ans.end());
      ans.erase(unique(ans.begin(), ans.end()), ans.end());
      t = ans.size();
      cout << min(t, c) << " ";
      for (i = 0; i < min(t, c); i++) {
        cout << ans[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}